Your setup would be ideal if you were standing on the equator at either the spring or fall equinox. The reason your result is lower than the actual is due to the geometry that results from the tilt of the Earth at that time of year and of your latitude on Earth. If you do this experiment on the 1st day of spring or 1st day of fall, then you'd only need to adjust due to your latitude. If you divide by the cosine of your latitude (you sound British so lets say South England ~50 deg), you'd get 6222 km (4000 / cos(50)).
@@OVAstronomyAn interesting method! I was surprised latitude didn't come into the equation. Or the angle at which the sun sets (You're assuming it goes down vertically) from which you can calculate how long it would take if it was vertical. Also I wondered about the t-squared outside the brackets. Shouldn't it be just in the theta-squared term? I sometimes see 2 or 3 sunsets when driving home from Mum's. Sun sets at Mum's house. I go up the road and it rises again - sets behind the hill, and sometimes pops up in a gap between hills just before it disappears for good! Perhaps seeing it set behind a hill 5-10 degrees above the true horizon is a good thing because refraction is less? PS if it's setting over the sea you might catch a glimpse of a green flash just as it disappears. I saw it maybe 2 or 3 times on holiday.
@@yahccs1 Those are some good additions to the simple model I presented! Perhaps in the future I'll recreate this video and take into account all these possible ideas you and others have suggested, so thank you! And yes, the t-squared term could be brought into the brackets for theta squared - the reason I brought it out was because I wanted to explicitly show that the rest of the theta squared term was one big constant, i.e. t is the only variable I was considering in theta, so I brought it out of the brackets, I should have made that more clear. Also it's cool that you mentioned the green flash over certain sunsets as I had a chance to look into and do some research into the phenomenon in my most recent year at university, so awesome that you mentioned it!
@@OVAstronomy Thanks for replying, I'm glad it was helpful. Your video inspired me to spend ages trying to work out how to find theta if you don't know time or radius... or the distance to the point on the horizon that the top of the sun just appears over. It can get quite complicated. Basically I wanted to work out the formula the other way around - knowing the radius approximately and working out how long the sun should take between those two sunsets, so how accurately you need to measure the time to get a reasonably accurate estimate of the radius! The 'flat' horizon is more than 90 degrees from the zenith so if you knew the angle over 90degrees that it is, you could work out the radius but that angle is so small and the difference between it and its slightly altered version between lying down and standing up is so much smaller, not really mesurable! Also the problem with the angle the sun goes down which I thought is the latitude angle (from vertical) but it's more than that if the sun sets over a point of higher altitude than the observer like over the hill. Still I thought the latitude angle would be good enough approximation for a sunset about 5 to 10 degrees higher than the 'flat horizon'. I'm still working on it...
Nice video and idea. Let me challenge our minds with this question. What if we don’t have modern cameras or time keepers other than an evenly scaled rope and a good night sky? In other worlds how did ancient philosopher figure out radius of and distance to earth moon sun?
Was the observation made over water? That can cause superior/inferior mirages - and the amount of deflection of the Sun’s position will vary based on how close your view is to the water
Yes, the observations were mostly made looking over the sea at the horizon. Mirages would definitely affect the apparent position of the sun - something worth taking into account in any future observations, if only sunglasses could completely remove them!
I think the problem with your calculations is that you weren't on the equator when you took the measurements. Therefore your circle arc had a smaller radius, and the one from google. Think about how much smaller the latitude circles are on a globe through the uk compared to the equator of the globe.
Great video! I’m surprised you haven’t had droves of flat earthers screaming “nuh-uh!” Walter Bisllin has an excellent discussion of refraction on his website, including a standard correction that can be applied for typical atmospheric conditions. I was also wondering if you could reduce error by starting when you see the start of sunset on the beach, then run up to an elevated position like a beach hotel room or observation deck. As long as you can get a good estimate of the elevation, you would have a bigger “h” and a longer time.
Thank you so much! I will definitely be checking out that correction. And yes, provided you can quickly get up to the higher observation point before the 2nd sunset. This would ever so slightly reduce the overall error by decreasing the strength of the uncertainty errors on height and time!
How can you work all this out knowing just one angle. Ie your 90 degree angle. You must know the radius in which case you know the circumference, or the distance to the Sun. Sure you know the time it takes but that does not give you the speed.
Radii meet tangents at right angles, so you know there's a 90 degree angle. The angle between the two sunsets is worked out through knowing how fast Earth rotates ie we know we turn 360 degrees approximately in 24 hours so we can work out how much the Earth has turned in any time period. Since we know our height we can write one length as Earth's radius r and the other as that plus your height (r+h). Because now we know 2 angles, one in terms of time and have written the lengths in terms of just one variable (r) , we can use trigonometry to obtain an expression for r in terms of time t. No need to already know the circumference or distance to the sun.
@@OVAstronomy To me you take a lot for granted. Is this how it was worked out 2,000 years ago and remember at that time the Sun moved around the Earth.
c'est un exercice classique de trigonométrie qui permet effectivement de déterminer la distance de 2 arbres installés sur une machine, ce n'est pas valable en astronomie, ce calcul suppose un "rayon de soleil" matériel qui n'existe pas, il n'est pas prouvé que le soleil nous envoie "des rayons de soleil" visible comme des lignes droites, d'autre part la terre et le soleil se déplace à grande vitesse ???. Un raisonnement "similaire" a été appliqué pour déterminer la distance de la terre à la lune , le chiffre de 300 milles kilomètres est contestable pour ne pas dire faux
Yes - there is a conversion error on screen when I show myself plugging in my height into the equation! A rookie mistake, I know!
I was about to ask if you'd ever built a model of Stonehenge for midgets.
Your setup would be ideal if you were standing on the equator at either the spring or fall equinox. The reason your result is lower than the actual is due to the geometry that results from the tilt of the Earth at that time of year and of your latitude on Earth. If you do this experiment on the 1st day of spring or 1st day of fall, then you'd only need to adjust due to your latitude. If you divide by the cosine of your latitude (you sound British so lets say South England ~50 deg), you'd get 6222 km (4000 / cos(50)).
I had almost forgotten about effects from Earth's tilt, so what you've said makes sense! Much appreciated.
@@OVAstronomyAn interesting method!
I was surprised latitude didn't come into the equation.
Or the angle at which the sun sets (You're assuming it goes down vertically) from which you can calculate how long it would take if it was vertical.
Also I wondered about the t-squared outside the brackets. Shouldn't it be just in the theta-squared term?
I sometimes see 2 or 3 sunsets when driving home from Mum's. Sun sets at Mum's house. I go up the road and it rises again - sets behind the hill, and sometimes pops up in a gap between hills just before it disappears for good!
Perhaps seeing it set behind a hill 5-10 degrees above the true horizon is a good thing because refraction is less?
PS if it's setting over the sea you might catch a glimpse of a green flash just as it disappears. I saw it maybe 2 or 3 times on holiday.
@@yahccs1 Those are some good additions to the simple model I presented! Perhaps in the future I'll recreate this video and take into account all these possible ideas you and others have suggested, so thank you! And yes, the t-squared term could be brought into the brackets for theta squared - the reason I brought it out was because I wanted to explicitly show that the rest of the theta squared term was one big constant, i.e. t is the only variable I was considering in theta, so I brought it out of the brackets, I should have made that more clear. Also it's cool that you mentioned the green flash over certain sunsets as I had a chance to look into and do some research into the phenomenon in my most recent year at university, so awesome that you mentioned it!
@@OVAstronomy Thanks for replying, I'm glad it was helpful. Your video inspired me to spend ages trying to work out how to find theta if you don't know time or radius... or the distance to the point on the horizon that the top of the sun just appears over. It can get quite complicated. Basically I wanted to work out the formula the other way around - knowing the radius approximately and working out how long the sun should take between those two sunsets, so how accurately you need to measure the time to get a reasonably accurate estimate of the radius! The 'flat' horizon is more than 90 degrees from the zenith so if you knew the angle over 90degrees that it is, you could work out the radius but that angle is so small and the difference between it and its slightly altered version between lying down and standing up is so much smaller, not really mesurable!
Also the problem with the angle the sun goes down which I thought is the latitude angle (from vertical) but it's more than that if the sun sets over a point of higher altitude than the observer like over the hill. Still I thought the latitude angle would be good enough approximation for a sunset about 5 to 10 degrees higher than the 'flat horizon'.
I'm still working on it...
Vernal and Autumnal are the proper terms.
Nice video and idea. Let me challenge our minds with this question.
What if we don’t have modern cameras or time keepers other than an evenly scaled rope and a good night sky? In other worlds how did ancient philosopher figure out radius of and distance to earth moon sun?
Was the observation made over water? That can cause superior/inferior mirages - and the amount of deflection of the Sun’s position will vary based on how close your view is to the water
Yes, the observations were mostly made looking over the sea at the horizon. Mirages would definitely affect the apparent position of the sun - something worth taking into account in any future observations, if only sunglasses could completely remove them!
I think the problem with your calculations is that you weren't on the equator when you took the measurements. Therefore your circle arc had a smaller radius, and the one from google. Think about how much smaller the latitude circles are on a globe through the uk compared to the equator of the globe.
Great video! I’m surprised you haven’t had droves of flat earthers screaming “nuh-uh!”
Walter Bisllin has an excellent discussion of refraction on his website, including a standard correction that can be applied for typical atmospheric conditions.
I was also wondering if you could reduce error by starting when you see the start of sunset on the beach, then run up to an elevated position like a beach hotel room or observation deck. As long as you can get a good estimate of the elevation, you would have a bigger “h” and a longer time.
Thank you so much! I will definitely be checking out that correction. And yes, provided you can quickly get up to the higher observation point before the 2nd sunset. This would ever so slightly reduce the overall error by decreasing the strength of the uncertainty errors on height and time!
@@OVAstronomy My pleasure, glad to help!
Wjay about light from the sun hitting the earth parallel?
More hints, the space has 360 days as one linear year which is only 1hr for the light while only 24hrs day for time
The minor differences in the metrics make little difference to the result, especially for someone who is only 179 cm tall.
Haha, well according to what I wrote on the screen I'm only 17.9 cm tall, so you could be correct 😅
Your calculation for angular speed uses a sidereal angle over a solar day.
How can you work all this out knowing just one angle. Ie your 90 degree angle. You must know the radius in which case you know the circumference, or the distance to the Sun. Sure you know the time it takes but that does not give you the speed.
Radii meet tangents at right angles, so you know there's a 90 degree angle. The angle between the two sunsets is worked out through knowing how fast Earth rotates ie we know we turn 360 degrees approximately in 24 hours so we can work out how much the Earth has turned in any time period. Since we know our height we can write one length as Earth's radius r and the other as that plus your height (r+h). Because now we know 2 angles, one in terms of time and have written the lengths in terms of just one variable (r) , we can use trigonometry to obtain an expression for r in terms of time t. No need to already know the circumference or distance to the sun.
@@OVAstronomy To me you take a lot for granted. Is this how it was worked out 2,000 years ago and remember at that time the Sun moved around the Earth.
c'est un exercice classique de trigonométrie qui permet effectivement de déterminer la distance de 2 arbres installés sur une machine, ce n'est pas valable en astronomie, ce calcul suppose un "rayon de soleil" matériel qui n'existe pas, il n'est pas prouvé que le soleil nous envoie "des rayons de soleil" visible comme des lignes droites, d'autre part la terre et le soleil se déplace à grande vitesse ???. Un raisonnement "similaire" a été appliqué pour déterminer la distance de la terre à la lune , le chiffre de 300 milles kilomètres est contestable pour ne pas dire faux
Actually, from new learning we know the Earth to be banana shaped
It's really concave bowl-shaped; it only appears flat.
@@comic4relief Witch!
Eratosthenes' method
opend your on eye... 🤔😋🧞♂️